Problem: If $x \dagger y = 2x^{2}-y^{2}$ and $x \star y = x^{2}+2y^{2}$, find $0 \dagger (2 \star 1)$.
Solution: First, find $2 \star 1$ $ 2 \star 1 = 2^{2}+2(1^{2})$ $ \hphantom{2 \star 1} = 6$ Now, find $0 \dagger 6$ $ 0 \dagger 6 = 2(0^{2})-6^{2}$ $ \hphantom{0 \dagger 6} = -36$.